(a) Two points X(32°N, 47°W) and Y(32°N, 25°E) are on the earth's surface. If it takes an aeroplane 11 hours to fly from X to Y along the parallel of latitu...

Answer Details

(a) Speed along the parallel of latitude

Both points are on latitude \(32^\circ\)N. The difference in longitude is

\[47^\circ + 25^\circ = 72^\circ\]

Distance along a parallel of latitude \(\phi\):

\[d = \frac{\theta}{360}\times 2\pi R\cos\phi\]

\[d = \frac{72}{360}\times 2\times\frac{22}{7}\times 6400\times\cos32^\circ\]

Using \(\cos32^\circ = 0.8480\):

\[d = 0.2\times \frac{44}{7}\times 6400\times 0.8480 \approx 6823\text{ km}\]

Speed \(= \dfrac{\text{distance}}{\text{time}} = \dfrac{6823}{11} \approx 620\) km/h.

Speed \(\approx 620\) km/h.

(b) Height of the kite

Let the foot of the vertical from the kite \(K\) meet the ground at \(O\). \(Q\) is nearer (elevation \(45^\circ\)) and \(P\) is \(15\) m further back (elevation \(35^\circ\)). Let \(OQ = d\) and height \(= h\).

From \(Q\): \(\tan45^\circ = \dfrac{h}{d} = 1 \Rightarrow d = h\).

From \(P\): \(\tan35^\circ = \dfrac{h}{d+15}\).

Substitute \(d = h\):

\[h = (h+15)\tan35^\circ = 0.7002(h+15)\]

\[h - 0.7002h = 10.503 \;\Rightarrow\; 0.2998h = 10.503 \;\Rightarrow\; h \approx 35\text{ m}\]

Height \(\approx 35\) m.